$\int^{\ln(8)}_{\ln(2)}\dfrac{e^x}{1+e^x}\,dx\, = $
Strategy Let's first find the indefinite integral $\int\dfrac{e^x}{1+e^x}\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{e^x}{1+e^x}\,dx\,$, we can use U-substitution. If we let $ {u=1+e^x}$, then ${du=e^x \, dx}$. So we have: $\begin{aligned}\int\dfrac{e^x}{1+e^x}\,dx\,&=\int\dfrac{1}{{1+e^x}}\,\cdot {e^x\, dx}\,\\\\\\\\ &=\int\dfrac{1}{ u}\, {du}\,\\\\\\\\ &=\ln|u|+C\\\\\\\\ &=\ln|1+e^x|+C\\\\\\\\ &=\ln(1+e^x)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{\ln(8)}_{\ln(2)}\dfrac{e^x}{1+e^x}\,dx\,&= \ln(1+e^x)\Bigg|^{\ln(8)}_{\ln(2)} \\\\\\\\ &= \ln\left(1+e^{\ln(8)}\right)-\ln\left(1+e^{\ln(2)}\right)\\\\\\\\ &=\ln\left(1+8\right)-\ln\left(1+2\right)\\\\\\\\ &=\ln(9)-\ln(3)\\\\\\\\ &=\ln\left(\dfrac{9}{3}\right)\\\\\\\\ &=\ln(3)\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{\ln(8)}_{\ln(2)}\dfrac{e^x}{1+e^x}\,dx\, = \ln3$